Question

4) Distance of line 4x – 3y = 9 from the point (3, 1) is
a) 9 units
b) -9 units
c) 9/5 units
d) 0 units​

Answer 1

Answer:

0 units..

Step-by-step explanation:

d=|Ax1+By1+C|/√A²+B². = FORMULA

d=|4×3+(-3)×1+(-9)|/√4²+(-3)²

d=|12-3-9|/√16+9

d=|12-12|/√25

d=0/5

d=0 units...

HOPE YOU HAVE UNDERSTOOD.

Answer 2

The distance of the line 4x – 3y = 9 from the point ( 3, 1 ) is (d) 0 units.

Given: The point ( 3, 1 ) and the line 4x – 3y = 9.

To Find: The distance of line 4x – 3y = 9 from the point ( 3, 1 ).

Solution:

We know that to find the distance of a point from a line, we make use of the formula,

           Distance = ( ax1 + by1 + c ) / √( a² + b² )                                 ....(1)

Where ( x1, y1 ) is the point from which the distance is to be measured on the line ax + by + c = 0.

Coming to the numerical, we are given;

The equation of the line is = 4x – 3y = 9

And the point is ≡ ( 3, 1 )

So, from (1), we can find the distance of the point from the given line;

              Distance = ( (4 × 3) - (3 × 1) - 9 ) / (√ 4²+ 3² )

                              = 0 / 5

                              = 0 units

So, the distance is zero units or we can say that the point ( 3, 1 ) lies on the line 4x – 3y = 9.

Hence, the distance of the line 4x – 3y = 9 from the point ( 3, 1 ) is (d) 0 units.

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